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Q1. a) Solve i. \( x^{2}+y^{A^{2}}=4 \) and \( y=2 \) ii. \( x^{2}+y^{7}=4 \) and \( y=x+1 \) b) what does your answer in (a) represent?

Ask by Osborne Welch. in Papua New Guinea
Mar 12,2025

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Answer

**Part (a):** **i.** Given the equations: \[ x^{2} + y^{A^{2}} = 4 \] \[ y = 2 \] Substituting \( y = 2 \) into the first equation: \[ x^{2} + (2)^{A^{2}} = 4 \] \[ x^{2} = 4 - 2^{A^{2}} \] \[ x = \pm \sqrt{4 - 2^{A^{2}}} \] **ii.** Given the equations: \[ x^{2} + y^{7} = 4 \] \[ y = x + 1 \] Substituting \( y = x + 1 \) into the first equation: \[ x^{2} + (x + 1)^{7} = 4 \] \[ x^{2} + x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 = 4 \] \[ x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 22x^{2} + 7x - 3 = 0 \] This is a 7th-degree polynomial equation, which is complex to solve analytically. Numerical methods or graphing techniques would be required to approximate the solutions. **Part (b):** The solutions to the system of equations in part (a) represent the points of intersection between the two curves defined by the equations. Specifically: - In part (a)i, the solutions \( x = \sqrt{4 - 2^{A^{2}}} \) and \( x = -\sqrt{4 - 2^{A^{2}}} \) indicate the points where the hyperbola-like curve \( x^{2} + y^{A^{2}} = 4 \) intersects the horizontal line \( y = 2 \). - In part (a)ii, the solutions to the 7th-degree polynomial equation represent the points where the 7th-degree curve \( x^{2} + y^{7} = 4 \) intersects the straight line \( y = x + 1 \). These intersection points are significant in various applications, such as determining equilibrium points in physics, solving optimization problems in economics, or analyzing geometric intersections in engineering.

Solution

Alright, I need to solve the given system of equations and then interpret the results. Let's tackle part (a) first. **Part (a):** **i. \( x^{2} + y^{A^{2}} = 4 \) and \( y = 2 \)** First, I'll substitute \( y = 2 \) into the first equation to find the value of \( x \). So, the equation becomes: \[ x^{2} + (2)^{A^{2}} = 4 \] Now, I need to solve for \( x \). However, there's a variable \( A \) in the exponent of \( y \), which complicates things. Since \( A \) is not defined, I'll assume it's a constant for the sake of solving. Let's denote \( A^{2} = k \), where \( k \) is a constant. So, the equation simplifies to: \[ x^{2} + 2^{k} = 4 \] Subtracting \( 2^{k} \) from both sides: \[ x^{2} = 4 - 2^{k} \] Taking the square root of both sides: \[ x = \sqrt{4 - 2^{k}} \] But since \( x \) can be positive or negative, the solutions are: \[ x = \sqrt{4 - 2^{k}} \quad \text{and} \quad x = -\sqrt{4 - 2^{k}} \] However, without a specific value for \( k \) (which depends on \( A \)), I can't provide numerical solutions. If \( A \) is given, I can substitute it back to find the exact values of \( x \). **ii. \( x^{2} + y^{7} = 4 \) and \( y = x + 1 \)** Again, I'll substitute \( y = x + 1 \) into the first equation to eliminate one variable. Substituting: \[ x^{2} + (x + 1)^{7} = 4 \] Now, I need to solve for \( x \). Expanding \( (x + 1)^{7} \) using the binomial theorem: \[ (x + 1)^{7} = x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 \] So, the equation becomes: \[ x^{2} + x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 21x^{2} + 7x + 1 = 4 \] Bringing all terms to one side: \[ x^{7} + 7x^{6} + 21x^{5} + 35x^{4} + 35x^{3} + 22x^{2} + 7x - 3 = 0 \] This is a 7th-degree polynomial equation, which is quite complex to solve analytically. Typically, such equations are solved numerically using methods like Newton-Raphson or graphing techniques to approximate the roots. Without specific tools or further information, I can't provide exact solutions here. **Part (b): What does your answer in (a) represent?** The solutions to the system of equations in part (a) represent the points of intersection between the two curves defined by the equations. Specifically: - In part (a)i, the equation \( x^{2} + y^{A^{2}} = 4 \) represents a hyperbola-like curve, and \( y = 2 \) is a horizontal line. The solutions \( x = \sqrt{4 - 2^{k}} \) and \( x = -\sqrt{4 - 2^{k}} \) indicate the points where the hyperbola intersects the horizontal line \( y = 2 \). - In part (a)ii, the equation \( x^{2} + y^{7} = 4 \) represents a curve that is more complex due to the exponent 7, and \( y = x + 1 \) is a straight line with a slope of 1 and a y-intercept of 1. The solutions to this system would represent the points where the 7th-degree curve intersects the straight line \( y = x + 1 \). These intersection points are significant in various applications, such as finding equilibrium points in physics, determining optimal solutions in economics, or solving geometric problems where two shapes intersect.

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The Deep Dive

To solve the equations, let's start with (i). Given \( y=2 \), substituting it into the first equation gives us \( x^{2}+2^{A^{2}}=4 \). This simplifies to \( x^{2}=4-2^{A^{2}} \). The solutions for \( x \) depend on the value of \( A \). For (ii), substituting \( y=x+1 \) into the second equation gives us \( x^{2}+(x+1)^{7}=4 \). This is a complex polynomial equation that may need numerical methods or graphical solutions to find values of \( x \). The answers in (a) represent points of intersection between curves defined by the equations. In geometry, these points often reveal solutions to systems of equations, showcasing where one equation's output fits into the context of another.

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